Can anyone me factor
as well as: →

For the first limit, you can factor the numerator as a difference of squares:

and the first factor can be factored the same way:

More generally, we have the formula

so really, for ,

In any case, we get

For the second limit, you can factor the denominator by grouping:

Then the limit is

2^y = ln 5, or   y(ln 2) = ln 5.   solving for y,   y = (ln 5) / (ln 2).   y=5

step-by-step explanation:

yes

step-by-step explanation: